Factor completely. $1 +12x +36x^2=$
Answer: Both $1$ and $36x^2$ are perfect squares, since $1=({1})^2$ and $36x^2=({6x})^2$. Additionally, $12x$ is twice the product of the roots of $1$ and $36x^2$, since $12x=2({1})({6x})$. $1x^2+12x+36 = ({1})^2 + 2({1})({6x})+({6x})^2$ So we can use the square of a sum pattern to factor: ${a}^2 + 2( a)( b)+ {b}^2 =({a} + {b})^2$ In this case, ${a}={1}$ and ${b}={6x}$ : $ ({1})^2 + 2({1})({6x})+({6x})^2 =({1} + {6x})^2$ In conclusion, $1 +12x +36x^2=(1 +6x)^2$ Remember that you can always check your factorization by expanding it.